Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
    public static void main(String args[]) throws IOException{
        process();
    }
    public static void process() throws IOException{
        BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(in.readLine());
        double arr[]=new double[n+1];
        StringTokenizer st=new StringTokenizer(in.readLine());
        for(int i=1;i<=n;i++)
            arr[i]=Double.parseDouble(st.nextToken());
        double dp[][]=new double[n+1][n+1];
        dp[0][0]=1.0;
        for(int i=1;i<=n;i++){
            for(int j=0;j<=i;j++){
                if(j==0)
                    dp[i][j]=dp[i-1][j]*(1.0-arr[i]);
                else
                    dp[i][j]=dp[i-1][j-1]*arr[i]+dp[i-1][j]*(1.0-arr[i]);
            }
        }
        double ans=0.0;
        for(int i=(n+1)/2;i<=n;i++)
            ans+=dp[n][i];
        System.out.println(ans);
    }
}